Question

Two charges are placed as shown in figure. Where should a third charge q be placed so that it remains at rest?

• A. $30$ cm from $9e$
• B. $40$ cm from $16$ e
• C. $40$ cm from $9e$
• D. $\left(1\right)or\left(2\right)$

Answer 1

The correct option is (C) $30$ cm from $9e$

Step 1,

Two charges are placed separated by the distance of 70 cm.

Let, the third charge q is placed at a distance x from $16e$ and hence its distance from $9e$ is $(70-x)cm$.

Step 2,

For the system to remain in equilibrium $F_1=F_2$

We can write,

or, $F_1=K\times \frac{9e\times q}{x^2}$

And

$F_2=\frac{K\times 16e\times q}{{(70-x)}^2}$

Equating $F_1$ and $F_2$ and we get,

Now,

$F_1=F_2$

or, $\frac{9e\times q}{x^2}=\frac{16e\times q}{{(70-x)}^2}$

or, $\frac{3}{x}=\frac{4}{{(70-x)}^2}$

or, $3(70-x)=4x$

or, $7x=48$

or, $210-3x=4x$

or, $x=\frac{210}{7}=30cm$

Hence the charge should be placed at 30 cm from the 9e.